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Links to specific sections in the text:
5.a. Wave function and Schrodinger equation for free
particles
5.b. The momentum operator and Schrodinger Equation for
particles
5.c. Particle in a box
5.d. Harmonic oscillator in Quantum mechanics
5.e. Operators and observables in quantum
mechanics
We have discussed the wave function for a free particle at a fixed time (t=0) before. Free particle here means that there are no external forces acting on the particle. We also saw that a time dependent monochromatic wave function has the form of
y(x,t) = A exp{ i ( kx - w t )},
where k and w satisfy the relation
k = 2 p p/h
w = 2 p E/h = 2 p p2 / 2mh
This wavefunction describes a particle that has a definite momentum. At the same time since |y(x,t) |2 = A2 is constant the particle has equal probability,
P(x) dx = |y(x,t) |2 dx = A2 dx
to be in every dx interval. In other words, the location of the particle, consistent with the Heisenberg uncertainty relation, is completely unknown.
To construct a wavefunction describing a particle with some information concerning its location we needed to construct a wave packet. This was done by taking a linear combination of the above type of wavefunctions with various values of k (or p, alternatively). such a wavefunction at t different from zero would be
y(x,t) = Ú exp{ i ( kx - w t )}f(k) dk,
or rather, after expressing the wave number by momentum we obtain
y(x,t) = (1/ h )1/2Ú exp{ i 2 p ( p x - E(p) t ) / h } f(p) dp,
where the notation E(p) = p2 / 2m was used. This is now the most general expression for a free particle wave packet. It provides a probability density at any given time
P(x) dx = |y(x,t) |2 dx
which can tell us what is the probability to find the particle between x and x+dx at time t. Wavefunctions multiplied by a coordinate independent constant provide the same probability distribution. Using this freedom we will usually use normalized wave functions, which satisfy the normalization condition,
Ú P(x) dx =Ú |y(x,t) |2 dx.
where the integral is taken over all x values.
The wavefunction
y(x,t) = (1/ h )1/2Ú exp{ i 2 p ( p x - E(p) t ) / h } f(p) dp,
is the most general form for a free particle. We will find a differential equation which this is the solution of. Conversely, all the solutions of the differential equations will be of this form, a well. The purpose of deriving such an equation is to be able to generalize it for the case when the particle is not free. After obtaining such an equation we will be able to tackle complicated dynamical problems, such as the properties of atoms, etc. To be able to discuss dynamical problems the equation to be derived should contain the time derivative of the wavefunction. Otherwise it is impossible to predict the time development of the system. Let us then take the time derivative of y(x,t). We obtain (after exchanging the order of integration over p and differentiation over t)
(d/dt)y(x,t)
= (1/ h )1/2Ú (- i 2p E(p)/h ) exp{ i 2 p ( p x - E(p) t ) / h } f(p) dp
= (1/ h )1/2Ú (- i 2p p2 /2 m h )
exp{ i 2 p ( p x - E(p) t ) / h } f(p) dp
In fact the "operator" that generates just a multiplier of E(p) under the integral is nothing else but
i (h/2 p ) (d/dt) -> E(p)op
This operator will be called the energy operator. Notice that, up to trivial constants, the second power of the momentum appeared in the integrand as a multiplier. There is another way to generate such a multiplier under the integral. Differentiating with respect to x twice also generates a multiplier of p2 up to some trivial constants
(d2/dx2)y(x,t)
= (1/ h )1/2Ú (- 4p2 p2 / h2 ) exp{ i 2 p ( p x - E(p) t ) / h } f(p) dp
Then we can combine together the this equation with that for the time derivative to obtain
i(h/2 p ) (d/dt)y(x,t) = - (h2 / 8p2 m)(d2/dx2)y(x,t),
which is the required differential equation for the wavefunction. Notice first that the imaginary unit, i, appears in the equation explicitly. This means that, in general, the wavefunction is bound to be complex. Secondly, we notice that an "operator" generating a multiplier p under the integral can also be defined easily. This is called the momentum operator.
-i (h/2 p ) (d/dx) -> pop.
Then the equation we obtained can easily written in terms of the energy and momentum operators:
Eop y(x,t) = ( pop2 / 2 m ) y(x,t) .
Never forget that these operators are defined through the differentiations. While we need operators to generate momentum multipliers under the integral defining the wave function, we need no such thing to obtain a multiplier depending on the coordinate only. In fact we can write down a natural generalization of the "operator equation" for particles that are not free, but rather are in a potential field:
Eop y(x,t) = ( pop2 / 2 m ) y(x,t) + V(x) y(x,t).
Now we can regard V(x) as an operator as well, but instead of differentiations it acts on the wavefunction simply by multiplying it with the function V(x). We can write out now the differential equation, the Schrodinger equation (in one dimension) for the wavefunction:
i(h/2 p ) (d/dt)y(x,t) = - (h2 / 8p2 m)(d2/dx2)y(x,t) + V(x) y(x,t).
This is not a derivationof the schrodinger equation. There is no derivation of the equation for nonzero potential. At the same time its form is made very plausible by the above considerations. This is the Schrodinger equation for one dimensional motion of a single particle only. It can be extended to three dimensions and to an arbitrary number of particles, so that for example
When one solves the Schrodinger equation for particles in a potential field then the simple form of the wave function is not valid, because the relation between momentum and energy E(p) = p2 /2m is lost. In other words, (infinitely) many different momenta are allowed even if the energy is fixed. In fact, solution of the Schrodinger equation corresponding to a fixed energy are very important. Bound states are such states. In bound states neither the momentum nor the coordinate are fixed, bu the energy is. We know that the states of the hydrogen atom correspond to definite energies. We call such states the eigenstates (German word: meaning its own) of the nergy operator. Imagine that in the Fourier decomposition of the wavefunction there is only one energy present. Then the time dependence of the wave function becomes very simple:
y(x,t) = exp{-i 2 pE t / h }y(x ),
where the last multiplier depends on x only. A state described by such a wave function is called a stationary state. It is almost independent of time, its time dependence is a perodic function multiplier. The time and coordinate dependences of this wavefunction factorize. It is easy to show that conversely, if we only assume that the wavefunction factorizes into y(x,t) = g(t)y(x ), then it is automatically of the above form, i.e. it is periodic in time.
These kind of wavefunctions satisfy
Eop y(x,t) = E y(x,t).
They are called the eigenstates of the energy operator, meaning that when the energy operator is applied to a wavefunction representing such a state then the action of the enrgy operator is the same as a multiplication by a constant, E. Substituting into the Schrodinger equation one obtains
Eop y(x,t) = E y(x,t) = ( pop2 / 2 m ) y(x,t) + V(x) y(x,t).
In other words, we obtain
E y(x,t) = - (h2 / 8p2 m)(d2/dx2)y(x,t) + V(x) y(x,t).
Now owing to the factorization of the wavefunction and that no derivative with respect to t appears any more we can multiply the equation by the factor
exp{i 2 pE t / h }, which cancels the time dependent factor from every term. Note that the differentiation with respect to x can be exchanged with a t dependent factor, which is a constant, as far as differentiation with respect to x goes. Then finally we obtain the equation for the wavefunction of a stationary state;
E y(x) = - (h2 / 8p2 m)(d2/dx2)y(x) + V(x) y(x).
This is the equation (or its generalization to three dimensions) that we will solve for various external potentials. Solving this equation will provide not only the wavefunction of the stationary states, but the energy spectrum as well. The energy spectrum will be a set of discrete energies that the system can take.
Finally a word about general properties of the wavefunction of a stationary state. Expressing the second derivative of the wavefunction from the Schrodinger equation we obtain
y"(x) = (8p2 m / h2 )[V(x) - E ]y(x).
The wavefunction, the absolute square of which is probability density, should be bounded. Then everywhere where the potential is bounded the second derivative of the wavefunction is bounded, as well. The indefinite integral of a function that is bounded is continuous. Thus the first derivative, and then, of course, the wavefunction itself are continuous. Everywhere where the potential is infinite the wavefunction must vanish, according to the equation, because V(x) y(x) must stay finite. It is a little trickier to find out what happens at points where the potential changes from infinite to finite. Applying appropriate limiting procedure one can prove that though the wavefunction is continuous (i.e. vanishes) at such points, but the derivative, though it stays bounded, may have a finite jump. Conditions like this are called boundary conditions.
The simplest bound state problem is the infinite potential well. The potential is defined by
V(x) = + infinity if x<0, or X>L,
V(x) = 0 if 0<x<L.
The wave function must vanish identically in a region where the potential is infinite. Thus, the wave function, being continuous, must also vanish at x=0 and x=L. In the region 0<x<L the Schrodinger equation, V(x) being zero, has the form
f"(x) = (8 p2 m / h2 )(-E)f(x)
Sice the potential energy is zero and the kinetic energy is always positive E>0. Then using the notation
k2 = 8 p2 m E/ h2 or k = (2p/ h )(2mE)1/2
where k is indeed the wave number, we have the simple differential equation for the wavefunction
f"(x) = - k2 f(x).
must stress that at this moment k2 is not known. It is a positive constant.
The general solution of this differential equation is
f(x) = A sin kx + B cos kx.
Now the wavefunction is also supposed to satisfy the boundary condition at x=0 and x=L. It must be continuous. Since it vanishes at x<0 and at x>L it also vanishes at x=0 and x=L. At x=0 we obtain
f(0) = A sin k0 + B cos k0 = B = 0.
Then we are left with the first term only and that term taken at x=L gives
f(L) = A sin kL = 0.
Now A cannot vanish, because that would make the wavefunction vanish identically. Thus, we must require that
sin kL =0.
This can be satisfied only if kL = np, or equivalently k = n p/ L, where n is an integer. In fact, there is no need to consider n=0. At n=0 k=0, as well and then f(x) = A sin kx = 0. Also there is no need to consider negative integers, because if n-> -n then k -> -k and the wavefunction does not really change because sin(-kx)= - sin kx. Consequently we only need to consider positive integers. For each positive integer n, n=1,2,... we obtain a different wavefunction and a different energy. The energies are
En = k2 h2 / 8 p2m = n2 h2 / 8 L2 m,
and the corresponding wavefunctions are
fn(x) = A sin (n px / L).
In other words there are infinitely many discrete energy levels. This can be contrasted with the similar problem in classical physics, in which all positive energies are admissible. In quantum physics we say that the energy levels are quantized. the first three energy levels and wavefunctions have the following form:
There are infinitely many energylevels rising to infinitely high energy. The nth wavefunction has n-1 nodes (zeros, aside from those at x=0 and x=L). These wavefunctions still have an undetermined constant, A. A can be determined from the requirement that |fn(x)|2dx represents the probability that the particle is between x and x+dx. This requies that the integral between 0 and x,
Ú |fn(x)|2dx = 1.
It is obvious that one can always choose A such that the above relation is satisfied. We obtain
1 = A2 Ú dx sin2(n px / L) = (A2 / 2) Ú dx [ 1 - cos(2 n px / L) ]
= A2 L / 2 - (A2 L / 4np ) sin(2 n px / L)|L0 = A2 L / 2,
because the last term vanishes at both limits. Then we can express A as
A = (2 / L)1/2
Then the normalized wavefunctions can be written as
fn(x) = (2 / L)1/2 sin (n px / L).
The absolute squares of these wavefunctions satisfy the standard probability interpretation.
Problem: What is the probability that the particle is in the middle half of the interval 0<x<L?
Solution: The probability, P, is given by the integral
Ú |fn(x)|2dx, where the limits are L/4 and 3L/4. We obtain, as before
P = 2/L Ú dx sin2(n px / L) = (1 / L) Ú dx [ 1 - cos(2 n px / L) ]
= 1 / 2 - (1 / 2np ) sin(2 n px / L)|3L/4L/4
= 1/ 2- (1 / 2np )[ sin(3 n p/ 2) - sin( n p/ 2)]
= 1 / 2 if n=even,
=1 /2 + 1/ np, if n = odd and of the form n=4k+1, k=0,1,... and
=1 / 2 - 1 / np, if n = odd and of the form n=4k+3, k=0,1,...
Problem: what is the expectation value of x?
Solution: <x> is defined as (wih integration imits 0 and L)
<x> = Ú x |fn(x)|2dx = (1/ L) Ú dx x [ 1 - cos(2 n px / L) ]
= (1/ L) x2/2|L0 = L / 2.
This is true because the second term can be calculated after integration by parts. x is differentiated and cos(2 n px / L) is integrated. The derivative of x is 1 and the intagral of cos(2 n px / L) is (L/ 2 n p ) sin(2 n px / L), which vanishes at both limit. The remaining integral over (L/ 2 n p ) sin(2 n px / L) gives
(L/ 2 n p )2 cos(2 n px / L) which is equal at the upper and lower limits and cancels. in other words for every one of the states, for all n, the average of x is just L/2, is at the middle of the interval. This makes sense because the square of each one of these wavefunctions is symmetric to the center point of the intervals.
Let us also discuss the slightly more complicated case of a potetial well of finite depth. We will get acquainted with another classically forbidden phenomenon, the penetration of a quantum particle into a classically forbidden region. Intuitively, one expects the existence of only a finite number of bound states, because in the case of the infinite potential bound states for large quantum number n had very high kinetic energies. If the potential well has a finite depth kinetic energies larger than the depth of the potential well are not possible, because particles heving such energies would escape. Define the potential now as V(x) =0 if |x|>L/2 and V(x) = - V0 if |x|<L/2. Again we need to solve the Schrodinger equation separately in the different regions, due to the different analytic form of the potential. After solving the equation in all three regions (x< - L/2, -L/2<x<L/2, and x>L/2) we must match the colutions at the boundaries. Since the potential is bounded both the wavefunction and its first derivative must go through the boundary continuously.
First leet us look at a sketch of the potential:
This potential is even, meaning V(-x) = V(x). This is obvious if we look at the above sketch of the potential: a function is even is it is the mirror image of itself when reflected to the vertical axis.
In general, if the potential is even then one can choose the wavefunctions to be either even or odd. This is easy to see. Take the Schrodinger equation
E y(x) = - (h2 / 8p2 m)(d2/dx2)y(x) + V(x) y(x)
assume that V(x) is even and substitute x -> -x. Then we obtain
E y(-x) = - (h2 / 8p2 m)(d2/dx2)y(-x) + V(x) y(-x).
This is true because the second derivative, being quadratic in dx does not change sign. This equation states that y(-x) is the solution of the same Schrodinger equation. Since the equation is linear, linear combination of solutions is a solution as well. Thus, y1(x) = y(x) + y(-x) and y2(x) = y(x) - y(-x). I is easy to see that y1(-x) = y1(x), so y1(x) is even and y2(-x) = - y2(x), which means that y2(x) is odd. If the potential is even than the eigenstates of the energy operator are always even or or odd. In fact, one can show that the ground state wavefunction is always even.
If the ground state wavefunction is even it is sufficient to solve the Schrodinger equation for x>0. If x<L/2, then the Schrodinger equation is
f"(x) = (8 p2 m / h2 )(-V0-E)f(x)
Since the free particle spectrum starts at E=0, we need to have E = K +V < 0 to have a bound state. But also K=E-V>0. Since V= -V0 we have E+V0 >0. Consequently we can write
f"(x) = - k2 f(x)
k2 = 8 p2 m (E + V0 )/ h2 or k = (2p/ h )[2m(E + V0 )]1/2
This equation has the same form as the equation for the infinite potential well except the definition of k2 is slightly different. Its general solution is
f(x) = A cos kx + B sin kx.
Now we stipulated that the solution is even, thus it must have the form
f(x) = A cos kx.
We can immediately see that the wavefunction cannot be zero inside the clasically forbidden domain of x>L/2. Suppose we choose k in such a way that the wavefunction vanishes at x=L/2. But then the derivative of the cos would not vanish at the same point. Since the derivative goes through the point x=L/2 continuously, as well, even if the function vanishes at x=L/2 it must be different from 0 in the immediate neighborhood of x=L/2.
In fact it is easy to find the wavefunction at x>L/2. There V(x) = 0. Since, as we remarked earlier, E<0. we have
f"(x) = k2 f(x),
where
k2 = 8 p2 m | E |/ h2 or k = (2p/ h )[2m | E |]1/2
The solution of the differential equation has a different form now (we assume k>0).
f(x) = C e-kx + D ekx
Let us see how this wavefunction behaves at large x. The first term decreases fast and practically zero for x>>1/k. The second term, however increases exponentially if x increases. Obviously, the integral over the square of a wavefunction having nonzero D would not be finite. Such a wavefunction could not be given probability inerpretation. Even if we disregard the condition that the square of the wavefunction is integrable, we would have a phyisically unacceptable solution that predicts finding the bound particle far away form the potential well is more probable than to find it in the well. So we can conclude that we must have D=0. In other words, we now know that the wavefunction is
f(x) = A cos kx if x<L/2 and
f(x) = C e-kx if x> L/2.
Continuity requires that
A cos(kL/2) = Ce-kL/2
while the continuity of the derivative requires that
- k A sin(kL/2) = - k Ce-kL/2
Dividing these two equations with each other gives
tan(kL/2) = k / k.
Remember now that k and k had a simple expression in terms of |E| to get
tan {(2p/ h )[2m(V0 - |E| )]1/2 } = [ | E | / ( V0 - |E| )]1/2
This equation cannot be solved analytically, but it can be solved graphically or numerically. It always has at least one solution for |E| that satisfies |E| < V0 E.g. for a shallow potential well when kL<<1, we can approximately write tan(kL/2) µ kL/2 and we get
(2p/ h )[2m(V0 - |E| )]1/2 = [ | E | / ( V0 - |E| )]1/2
Squaring this equation we obtain
(2p/ h )2 2m(V0 - |E| )2 = | E |,
a second order equation for |E|. The value one obtins is not really interesting but it is worth while to plot the wavefunction. In the next figure both the potential (ground state energy) and the groundwavefunction are plotted in the same diagram.
The stunning feature of the wavefunction is that it does not vanish inside the classically forbidden region but it penetrates it to a depth of x µ 1/k. This is a general quantum phenomenon. Particles can always penetrate classically forbidden regions, but the penetration decreases exponentially as we go deeper inside the classically forbidden region. The average depth of penetration is given exactly by 1/k.
Now imagine that we put two pieces of metal to each other, but assume that there is an oxide layer on the surface, so that they are separated by an insulator. This can be represented by a double potential well. Also assume that the metals are different so that the energy (Fermi level) of the highest energy electrons is different. Then the electrons from one metal can go through the potential barrier formed by the oxide layer into the other metal if that is energetically favorable. This phenomenon is called tunneling and will be studied later.
Until now we have been discussing artificially constructed rectangular potentials. In some cases these give a good approximation of physical situations, in some other cases when the physical potential is smooth they do not. Suppose we study a dynamical system near stable equilibrium. A system is in stable equilibrium if it is at the minimum of the potential energy, V(x), at x = x0 . If a system is near the equilibrium then | x- x0 | is small. Then the potential can be represented by its Taylor series at x = x0 . We can write
V(x) = V(x0 ) + (x-x0 ) V'(x0 ) + (1/2) (x-x0 )2 V"(x0 ) + ...
Since the energy scale is arbitrary, we can always define it so that V(x0 ) = 0. We can also define the coordinate system such that x0 = 0. Furthermore, since the point x0 = 0 is the minimum of the potential, at this point V'(x0 ) = 0, as well. We can conclude that near the equilibrium the potential energy can be written as
V(x) = (1/2) x2 V"(0 ) = (m w2 / 2 ) x2
This is nothing else but the potential of the harmonic oscillator. This is why the harmonic oscillator is of such a central importance in physics. Every potential can be approximated by a harmonic oscillator potential near equilibrium. Therefore it is important to investigate the harmonic oscillator potential in quantum mechanics. The quantum mechanical is defined by the Schrodinger equation
E y(x) = - (h2 / 8p2 m)y"(x) + (m w2 / 2 ) x2 y(x)
= ( p2 / 2 m + m w2 x2 / 2 ) y(x)
where p means the momentum operator p = - i h / 2p d/ dx. Solving the Quantum mechanical problem, just like in the case of the infinite potential well (particle in a box) consists of finding the energy eigenvalues and eigenfunctions. Since the potential is an even function of x, we know that the eigenfunctions will be either even or odd. The potential has the form of a parabola. The walls of this potential, just like that of the infinite potential well, rise up infinitely high. An obvious guess that there are infinitely many energy levels rising infinitely high. To get an idea concerning the solution, consider the differential equation at very large x. No matter how large x is, at very large x E<<m w2 x2 / 2. Then the differential equation has the form
y"(x) µ 4 m2 w2 p2 x2 / h2 y(x).
This can be solved, again approximately, at large x if we require that
y'(x) µ ± 2 m w p x / h y(x).
This equation can be solved exactly. The solution is easily obtained if we divide by y(x) and obtain
y'(x) / y(x) = d logy(x) / dx = ± 2 m w p x / h
Integrating this equation we obtain
logy(x) = const. ± m w p x2 / h
or y(x) =A exp{ ± m w p x2 / h }
Now the solution with the positive sign in the exponent is inadmissable. It would imply hugely increasing probabilities when the particle is far in the classically forbidden region. So we are bound to keep the negative sign in the exponent only. In fact, this is an exact solution of the Schrodinger equation, not only an approximate one. We get
y'(x ) = - (2 m w p x / h ) A exp{ - m w p x2 / h }
and
y"(x) = ( 4 m2 w2 p2 x2 / h2 ) A exp{ - m w p x2 / h } - 2 m w p / h ) A exp{ - m w p x2 / h }
= ( 4 m2 w2 p2 x2 / h2 - 2 m w p / h ) y(x).
Comparing with the original equation
E y(x) = - (h2 / 8p2 m)y"(x) + (m w2 / 2 ) x2 y(x)
= ( p2 / 2 m + m w2 x2 / 2 ) y(x)
we can see that provided we choose
E 8p2 m / h2 = 2 m w p / h,
or equivalently,
E = w h / 4p
the equation is exactly satisfied. Thus the ground state energy is E = w h / 4p = fh/2. It is exactly one half of the energy of a photon of frequency f. The eigenfunction is a Gaussian function, which vanishes very fast when |x| is large, but it is nowhere exactly zero. In fact, at points where the potential energy V(x) = m w2 x2 / 2 > E = w h / 4p the particle is classically forbidden to exist. Solving the equation for x, we obtain these points as x = ±[ h / 2pm w]1/2 = ± x0 Still, quantum mechanically, there is a nonvanishing probability that the particle can be found in such a region. We could reprresent the energy level and the ground state wavefunction of the harmonic oscillator as
Let us discuss now excited states. The asymptotic behavior of the wavefunction, as derived above, should be the same irrespective of the energy of the state. Then it is the same for all excited states. This means that to get excited state wave functions we need to multiply the ground state wavefunctionbe a factor that varies slower than the Gaussian representing the ground state wavefunction, so that it would not spoil the asymptotic behavior. It is fairly easy to show that only polynomial can be used for this purpose. Since the potential is an even function of x, only even or odd wavefunctions should be considered. The Gaussian is an even function. An even function multiplied by another function (i.e. polynomial) is even or odd if the polynomial is even or odd. A polynomial is even if it contains even powers of x only. It is odd if it contains odd powers of x only. Therefore, one can attempt to use the simplest nontrivial polynomial, P1(x) = x, which is odd. In other words one can try the wavefunction
y1(x) = C x exp{ - m w p x2 / h }.
Taking derivatives we obtain
dy1(x)/ dx= C exp{ - m w p x2 / h }
-( C m w p x2 / 2h ) exp{ - m w p x2 / h }
d2 y1(x) / dx2= - 3 ( C m w p x / 2h ) exp{ - m w p x2 / h }
+ ( C m2 w2 p2 x3 / h2 ) exp{ - m w p x2 / h }
=-3 ( m w p / 2h ) y1(x) + ( m2 w2 p2 x2 / h2 ) y1(x)
Comparing with the equation
d2 y1(x) / dx2= (h2 / 8m p2 ) y1(x) [ V(x) - E]
we can see that as expected from the correct asymptotic behavior the x2 term is correct and we obtain
E1 =3 hw / 4p = 3 hf /2.
The excitation energy is hf, the same as the energy of a Planck quantum. In fact, one can show that using higher and higher order polynomials, the so called Hermite polynomials, we can write the nth energy eigenfunction as
yn(x) = Hn (x )exp{ - m w p x2 / h },
where Hn (x ) is an nth order (alternatively even en odd) polynmial. The corresponding energies are
En =(n+1/2) hw / 2p = (n+1/2) hf .
We can view the energy spectrum as follows. There is a ground state (or as sometimes called zero point) energy, hf /2. The system cannot go lower than that (a heuristic explanation why that cannot happen will be given later). Then we can give energy to the system in chunks of hf. We can add only an integer number of energy quanta, that is how the quantum system differs from a classical system. Remember this is exactly what Planck assumed in his derivation of the Planck law. The oscillators that make up the enery of the cavity (which can be regarded as a infinite potential well) can only have discrete energies. Based on quantum mechanics, we proved the basic postulate of Planck that allowed to derive his law.
Let us now return to the zero point energy (or ground state energy). Is there a general physical argument why it isnot allowed for the quantum mechanical system to have zero energy? The answer is simple, based on the uncertainty principle. If the energy is zero then the momentum is zero as well. If we know that the momentum is zero then the uncertainty of the momentum is zero as well. Then the uncertainty of the coordinate should be infinite. But in the case of the harmonic oscillator if the energy would be zero then particle would be sitting at the bottom of the potential well and as such the uncertainty of the coordinate would be zero. There is a clear contradiction with the uncertainty principle. In fact we can estimate the ground state energy using the uncertainty principle.
Problem: Calculate the expectation value of the energy in the ground state of the harmonic oscillator using the uncertainty principle. Use the fact that <x> = <p> = 0.
Solution: The energy is
E = p2 /2m + m w2 x2/ 2
In an energy eigenstate the energy is exactly known so then E = <E>. But
<E> =< p2 > /2m + m w2 < x2> / 2 = < (p-<p>)2 > /2m + m w2 < (x-<x>)2> / 2 ,
where we used <x> = <p> = 0. But then this is nothing else but
E = Dp2 /2m + m w2 Dx2/ 2
Now using the uncertainty relation Dp2 > h2 /( Dx28p2 ) we obtain
E > h2 /( 2mDx28p2 ) + m w2 Dx2/ 2
The right hand side is a function of Dx2 . It is always larger than its minimum, as a function of Dx2 . To find this minimum we need to differentiate it with respect to Dx and set the derivative equal to zero. We obtain
0 = - h2 /( mDx38p2 ) + m w2 Dx
Solving this equation we obtain
Dx = [h2 /( m2Dx38p2 w2 )]1/4
Substituting Dx back into the expression of energy we get the value of the energy at the minimum
E = hbar w / 2
which is the exact ground state energy. The exact ground state energy is obtained, because the uncertainty relation was saturated by the solution that is a Gaussian wave function.
Certain quantities are measurable in quantum mechanics. A quantum mechanical state is defined by the wave function. In a certain state some quantities have well defined (sharp) values, some other quantities have smeared values when only the probability of the outcome of a measurment can be predicted. Example: The plane wave is described by the wavefunction
y(x,t) = ei(px-Et)/hbar
This wave function describes a free particle of momentum p. In other words, when the particle is in the state described by this wave function its momentum is sharp and has value p. The coordinate of the particle is, however, completely indetereminate. The measurment of the coordinate can lead to any value with the same probability. If the uncertainty of a physical variable, observable, is zero then the physical variable is sharp.
In the Gaussian wave packet neither the momentum nor the coordinate are sharp. They are both smeared and there is finite probability to find the particle in a finite region. It also can have momenta in a certain range.
As we learned earlier the wave function has a probability interpretation. The probability of finding the particle in the interval (x, x+dx) is | y(x,t)|2 dx. Thus the average of x is
<x> = Ú x | y(x,t)|2 dx = Ú y*(x,t) x y(x,t) dx
We can view this as applying the operator of coordinate (that is just a multiplication by the coordinate) to the wave function y(x,t) and then multiplying this by the complex conjugate of the wave function, y*(x,t), and integratibg over x. In fact, we can generalize this method to physical variables other then the coordinate, the operators of which are not as simple. An example is the momentum operator
<p> = Ú y*(x,t) pop. y(x,t) dx
= - i hbar Ú y*(x,t) (d y(x,t) / dx) dx
or to the energy operator
<E> = Ú y*(x,t) Eop. y(x,t) dx
= - i hbar Ú y*(x,t) (d y(x,t) / dt) dx
In a stationary state the wave function is the eigenfunction of the energy operator, with eigenvalue E, Eop. y(x,t) = E y(x,t), so we have
<E> = Ú y*(x,t) Eop. y(x,t) dx = E Ú y*(x,t) y(x,t) dx
= E
because the wave function is normalized,
Ú y*(x,t) y(x,t) dx= Ú |y(x,t)|2 dx = 1.
In other words the energy has a sharp value in a stationary state.
Defining the Hamilton operator as
H = K + V = - (hbar)2d2/dx2 + V(x)
we can calculate its expactation value in a stationary state and get the same result.
<H> = E
Here we use the fact that the time independent wave function is an eigenfunction of the Hamilton operator with eigenvalue E. This is just the content of the time independent Schrodinger equation
E y(x,t) = - (h2 / 8p2 m)(d2/dx2)y(x,t) + V(x) y(x,t).